A124780 a(n) = gcd(A(n), A(n+2)) where A(n) = A000522(n) = Sum_{k=0..n} n!/k!.

1, 2, 5, 2, 1, 2, 1, 10, 1, 2, 13, 2, 5, 2, 1, 2, 1, 10, 1, 2, 1, 2, 5, 26, 1, 2, 1, 10, 1, 2, 1, 2, 5, 2, 37, 2, 13, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 1, 26, 1, 2, 5, 2, 1, 2, 1, 10, 1, 2, 1, 2, 65, 2, 1, 2, 1, 10, 1, 2, 1, 74, 5, 2, 1, 26, 1, 10, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 13, 2, 1, 2, 5, 2, 1, 2, 1

Offset

0,2

Comments

a(n) divides n+3 because A(n+2) = (n+2)(n+1)*A(n) + n+3.

Links

Antti Karttunen, Table of n, a(n) for n = 0..4096

J. Sondow, A geometric proof that e is irrational and a new measure of its irrationality, Amer. Math. Monthly 113 (2006) 637-641.

J. Sondow, A geometric proof that e is irrational and a new measure of its irrationality, arXiv:0704.1282 [math.HO], 2007-2010.

J. Sondow and K. Schalm, Which partial sums of the Taylor series for e are convergents to e? (and a link to the primes 2, 5, 13, 37, 463), II, Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010.

Index entries for sequences related to factorial numbers

Formula

a(n) = gcd(A000522(n), A000522(n+2)) = (n+3)/A124782(n)

Example

a(2) = gcd(A(2), A(4)) = gcd(5, 65) = 5.

Mathematica

(A[n_] := Sum[n!/k!, {k, 0, n}]; Table[GCD[A[n], A[n+2]], {n, 0, 100}])
GCD[#[[1]], #[[3]]]&/@Partition[Table[Sum[n!/k!, {k, 0, n}], {n, 0, 100}], 3, 1] (* Harvey P. Dale, Jun 14 2022 *)

Prog

(PARI)
A000522(n) = sum(k=0, n, binomial(n, k)*k!); \\ This function from Joerg Arndt, Dec 14 2014
A124780(n) = gcd(A000522(n), A000522(n+2)); \\ Antti Karttunen, Jul 07 2017

Crossrefs

Cf. A000522, A093101, A123899, A123900, A123901, A124779, A124781, A124782.

Sequence in context: A342059 A061176 A180957 * A369872 A108437 A226029

Adjacent sequences: A124777 A124778 A124779 * A124781 A124782 A124783

Keyword

nonn

Author

Jonathan Sondow, Nov 07 2006

Status

approved